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Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a smooth transformation. Define the pullback $T^*: C^k (\mathbb{R}^m) \rightarrow C^k (\mathbb{R}^n)$ (With $C^k(\mathbb{R}^n)$ being the set of functionals on $k$-cells on $\mathbb{R}^n$) by $T^*: Y \mapsto Y \circ T$.

Thus, the pullback of $Y \in C^k (\mathbb{R}^m)$ is the functional on $C_k (\mathbb{R}^n)$ $$T^*Y:\phi \mapsto Y(T\circ \phi)$$

Why is the pullback linear, and why does $(T \circ S)^* = S^* \circ T^*$?

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Both facts follow from a direct verification: have you tried to simply use the definitions of everything in sight? – Mariano Suárez-Alvarez May 2 '12 at 1:23
I guess it is just difficult for me to understand how to correctly manipulate $k$-cells. – Sam Kidd May 2 '12 at 1:26
Say you want to check that $(T\circ S)^*=S^*\circ T^*$. The two sides of the equality are functions, so to chck equality you need to apply both sides to an element of their domain and use their definitions to compute: the two results must be equal. – Mariano Suárez-Alvarez May 2 '12 at 1:30
Alright, is it actually this easy then? $$(T \circ S)^* (Y) = Y \circ (T \circ S)$$ and $$S^* \circ T^*(Y) = S^* (Y \circ T) = Y \circ T \circ S$$ and to show linearity: $$T^*(aY + X) = (aY + X)(T) = aY(T) + X(T)$$ – Sam Kidd May 2 '12 at 1:34
Yes :) (Although in the part about linearity you are missing a few $\circ$s.) – Mariano Suárez-Alvarez May 2 '12 at 1:39
up vote 0 down vote accepted

Let $X,Y$ be arbitrary functionals (and $a$ an arbitrary constant). Since linearity is preserved on $k$-forms:

\begin{align} T^*(aY + X) &= (aY + X) \circ T \\ &= (aY + X) (T) \\ &= aY(T) + X(T) \\ &= aT^*(Y) + T^*(X). \end{align}

\begin{align} (T \circ S)^* (Y) &= Y \circ (T \circ S) \\ &= Y \circ T \circ S \\ &= S^*(Y \circ T) \\ &= S^* \circ T^* (Y). \end{align}

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